Given n, count all 'a' and 'b' that satisfy the condition a^3 + b^3 = n. Where (a, b) and (b, a) are considered two different pairs
Examples:
Input: n = 9
Output: 2
Explanation: 1^3 + 2^3 = 9 and 2^3 + 1^3 = 9
Input: n = 28
Output: 2
Explanation: 1^3 + 3^3 = 28 and 3^3 + 1^3 = 28
[Naive Approach] Using Nested Loops - O(n^2) time and O(1) space
The approach uses a nested loop to check all pairs (a, b) where a3+b3=n. If the condition holds, the count is incremented.
C++
#include <iostream>
using namespace std;
int countPairs(int n) {
int count = 0;
for (int a = 1; a <= n; a++) {
for (int b = 0; b <= n; b++) {
if (a * a * a + b * b * b == n) {
count++;
}
}
}
return count;
}
int main() {
int n = 9;
cout << countPairs(n) << endl;
return 0;
}
#include <stdio.h>
int countPairs(int n) {
int count = 0;
for (int a = 1; a <= n; a++) {
for (int b = 0; b <= n; b++) {
if (a * a * a + b * b * b == n) {
count++;
}
}
}
return count;
}
int main() {
int n = 9;
printf("%d\n", countPairs(n));
return 0;
}
/*package whatever //do not write package name here */
import java.io.*;
public class GfG {
static int countPairs(int n) {
int count = 0;
for (int a = 1; a <= n; a++) {
for (int b = 0; b <= n; b++) {
if (a * a * a + b * b * b == n) {
count++;
}
}
}
return count;
}
public static void main(String[] args) {
int n = 9;
System.out.println(countPairs(n));
}
}
def count_pairs(n):
count = 0
for a in range(1, n + 1):
for b in range(n + 1):
if a**3 + b**3 == n:
count += 1
return count
n = 9
print(count_pairs(n))
using System;
class GfG {
static int CountPairs(int n) {
int count = 0;
for (int a = 1; a <= n; a++) {
for (int b = 0; b <= n; b++) {
if (a * a * a + b * b * b == n) {
count++;
}
}
}
return count;
}
static void Main() {
int n = 9;
Console.WriteLine(CountPairs(n));
}
}
function countPairs(n) {
let count = 0;
for (let a = 1; a <= n; a++) {
for (let b = 0; b <= n; b++) {
if (a ** 3 + b ** 3 === n) {
count++;
}
}
}
return count;
}
let n = 9;
console.log(countPairs(n));
[Expected Approach] Finding different pairs - O(n1/3) time and O(1) space
While traversing numbers from 1 to the cube root of n, Subtract the cube of the current number from n and check if their difference is a perfect cube.
C++
// C++ program to count pairs whose sum cubes is n
#include<bits/stdc++.h>
using namespace std;
int countPairs(int n) {
int count = 0;
// Check for each number 1 to cbrt(n)
for (int i = 1; i <= cbrt(n); i++) {
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given N
int diff = n - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = cbrt(diff);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
return count;
}
int main() {
int n = 9;
cout << countPairs(n) <<"\n";
return 0;
}
#include <stdio.h>
#include <math.h>
int countPairs(int n) {
int count = 0;
// Check for each number 1 to cbrt(n)
for (int i = 1; i <= cbrt(n); i++) {
// Store cube of a number
int cb = i * i * i;
// Subtract the cube from given N
int diff = n - cb;
// Check if the difference is also a perfect cube
int cbrtDiff = cbrt(diff);
// If yes, then increment count
if (cbrtDiff * cbrtDiff * cbrtDiff == diff)
count++;
}
return count;
}
int main() {
int n = 9;
printf("%d\n", countPairs(n));
return 0;
}
// Java program to count pairs whose sum cubes is n
class GfG {
// method to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
static int countPairs(int n) {
int count = 0;
// Check for each number 1 to cbrt(n)
for (int i = 1; i <= Math.cbrt(n); i++) {
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given n
int diff = n - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = (int) Math.cbrt(diff);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
return count;
}
public static void main(String args[]) {
int n = 10;
System.out.println(countPairs(n));
}
}
# Python 3 program to count pairs whose sum cubes is n
import math
def countPairs(n):
count = 0
# Check for each number 1 to cbrt(n)
for i in range(1, int(math.pow(n, 1/3)) + 1):
# Store cube of a number
cb = i * i * i
# Subtract the cube from given n
diff = n - cb
# Check if the difference is also a perfect cube
cbrt_diff = round(diff ** (1/3))
# If yes, then increment count
if cbrt_diff * cbrt_diff * cbrt_diff == diff:
count += 1
return count
if __name__ == "__main__":
n = 2
print(countPairs(n))
// C# program to count pairs whose sum cubes is n
using System;
class GfG {
// method to count the pairs satisfying
// a ^ 3 + b ^ 3 = N
static int countPairs(int n) {
int count = 0;
// Check for each number 1 to cbrt(N)
for (int i = 1; i <= Math.Pow(n,(1.0/3.0)); i++) {
// Store cube of a number
int cb = i*i*i;
// Subtract the cube from given N
int diff = n - cb;
// Check if the difference is also
// a perfect cube
int cbrtDiff = (int) Math.Pow(diff,(1.0/3.0));
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
return count;
}
public static void Main() {
int n = 9;
Console.Write(countPairs(n));
}
}
// Javascript program to count pairs whose sum cubes is n
function countPairs(n) {
let count = 0;
// Check for each number 1 to cbrt(n)
for (let i = 1; i <= parseInt(Math.pow(n,(1.0/3.0)), 10); i++)
{
// Store cube of a number
let cb = i*i*i;
// Subtract the cube from given n
let diff = n - cb;
// Check if the difference is also
// a perfect cube
let cbrtDiff = parseInt(Math.pow(diff,(1.0/3.0)), 10);
// If yes, then increment count
if (cbrtDiff*cbrtDiff*cbrtDiff == diff)
count++;
}
// Return count
return count;
}
// driver code
const n = 9;
console.log(countPairs(n));
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